Question

If asin² (x) + bcos² (y) = c, bsin² (y) + acos² (y) = d and atan (x) = btan (y)
Then prove that : a²/b² = (d - a) (c - a)/(b - c) (b - d)

solve by eliminating x and y.​

Answer 1

Correct question :-

If asin² (x) + bcos² (x) = c, bsin² (y) + acos² (y) = d and atan (x) = btan (y). Then prove that : a²/b² = (d - a) (c - a)/(b - c) (b - d).

Given :-

$\sf a sin^2x+bcos^2x=c$

$\sf bsin^2y+acos^2y=d$

$\sf atan \ x= b tan \ y$

To Prove :-

$\sf \dfrac{a^2}{b^2}= \dfrac{(d-a)(c-a)}{(b-c)(b-d)}$

Solution :-

$\sf asin^2x+bcos^2x=c$

$\bf\dag \ sin^2x= 1- cos^2x$

$: \implies \sf \ a(1-cos^2x)+bcos^2x=c$

$: \implies \sf \ a-acos^2x+bcos^2x=c$

$: \implies \sf \ -acos^2x+bcos^2x= c-a$

$: \implies \sf \ cos^2x(b-a)= c-a$

$: \implies \sf \ cos^2x= \dfrac{c-a}{b-a} \ \ \ \ \ \ \ -(1)$

$\bf\dag \ cos^2x=1-sin^2x$

$: \implies \sf \ asin^2x+b(1-sin^2x) = c$

$: \implies \sf \ asin^2x+b-bsin^2x=c$

$: \implies \sf \ asin^2x-bsin^2x= c-b$

$: \implies \sf \ bsin^2x-asin^2x= b-c$

$: \implies \sf \ sin^2x(b-a) =b-c$

$: \implies \sf \ sin^2x= \dfrac{b-c}{b-a} \ \ \ \ \ \ \ \ \ -(2)$

Dividing eq(2) by eq(1) we get,

$: \implies \sf \ tan^2x= \dfrac{b-c}{c-a}$

$\sf bsin^2y+acos^2y = d$

$\bf\dag \ sin^2y= 1-cos^2y$

$: \implies \sf \ b(1-cos^2y)+acos^2y=d$

$: \implies \sf \ b-bcos^2y+acos^2y = d$

$: \implies \sf \ - bcos^2y+acos^2y=d-b$

$: \implies \sf \ bcos^2y-acos^2y=b-d$

$: \implies \sf \ cos^2y(b-a)=b-d$

$: \implies \sf \ cos^2y = \dfrac{b-d}{b-a} \ \ \ \ \ \ \ \ \ \ \ -(3)$

$\bf \dag \ cos^2y = 1-sin^2y$

$: \implies \sf \ bsin^2y+a(1-sin^2y) = d$

$: \implies \sf \ bsin^2y+a-asin^2y=d$

$: \implies \sf \ b sin^2y-asin^2y = d-a$

$: \implies \sf \ sin^2y(b-a)=d-a$

$: \implies \sf \ sin^2y = \dfrac{d-a}{b-a} \ \ \ \ \ \ \ \ -(4)$

Dividing eq(4) by eq(3) we get,

$: \implies \sf \ tan^2y = \dfrac{d-a}{b-d}$

Given that,

a tanx = b tany

$: \implies \sf \ \dfrac{a}{b} = \dfrac{tany}{tanx}$

By squaring both sides we get,

$: \implies \sf \ \dfrac{a^2}{b^2}= \dfrac{tan^2y}{tan^2x}$

Substituting the values of tan²x and tan²y,

$: \implies \sf \ \dfrac{a^2}{b^2} = \dfrac{(d-a)}{(d-b)} \times \dfrac{(c-a)}{(b-c)}$

$: \implies \sf \ \dfrac{a^2}{b^2}=\dfrac{(d-a)(c-a)}{(b-c)(b-d)}$

Hence proved.